# Thread: Exponential model from 2 points

1. ## Exponential model from 2 points

I'm given two points. The model has to look like $\displaystyle y=ae^bx$. I have to make a model that "fits the points shown in the table":

I started to do this one based off of a example I found in the book, but I just get lost in the beginning.

$\displaystyle 4=Ce^k(0)$
$\displaystyle 4/e^k(0)$
$\displaystyle (1/3)=(4/e^k(0))(e^k7)$

That's where I am (if I'm even on the right track).

2. Originally Posted by BeSweeet
I'm given two points. The model has to look like $\displaystyle y=ae^bx$. I have to make a model that "fits the points shown in the table":

I started to do this one based off of a example I found in the book, but I just get lost in the beginning.

$\displaystyle 4=Ce^k(0)$
$\displaystyle 4/e^k(0)$
$\displaystyle (1/3)=(4/e^k(0))(e^k7)$

That's where I am (if I'm even on the right track).
Do you mean $\displaystyle y = ae^{bx}$

If so:

$\displaystyle f(0) = a = 4$

$\displaystyle f(7) = 4e^{7b} = \frac{1}{3}$

$\displaystyle e^{7b} = \frac{1}{12}$

Solve for b.

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If $\displaystyle y = ae^bx$ then you can't solve with two unknowns and only one usable equation

3. After doing all that, I got (as my final answer):
$\displaystyle y=4e^(-3.55t)$

... Which isn't right. What am I doing wrong?

4. Originally Posted by BeSweeet
After doing all that, I got (as my final answer):
$\displaystyle y=4e^(-3.55t)$

... Which isn't right. What am I doing wrong?
Perhaps the exact value of b is required: $\displaystyle b = - \frac{\ln 12}{7}$.